Let $x_i \in Z$ , such that $|x_1| + |x_2| + \dots + |x_{10}| = 100$. Find number of solutions
I think the answer is in the form of an alegebric summation instead of a number, right? If yes then count the number of solutions in terms of K when K terms of out those ten are zero( you may get an idea from MathWiz's answer) and then add all the cases as k varies from 0 to 9. What would the next cases look like? Or how to find the solutions?
The amount of positive integer solutions would be $\frac{110!}{9! X 100 !} $, at home solution I could or could not put a minus sign in the number (because it's in module)
For a 10-double I would have $2 ^ {10}$. Am I right?
Case 1: All $x_{i}\neq 0$: There are $\binom{100-10+9}{9}=\binom{99}{9}$ ways to distribute the 100, since $x_{i}$ can be negative, you have $2^{10}$ ways to distribute the negative sign. So this gives you a total of $2^{10}\binom{99}{9}$ ways. Case 2: Only one out of $x_{i}$ is $0$: There are $\binom{10}{1}$ ways to make any one of $x_{i}$ equal to $0$ and $\binom{99}{8}$ ways to distribute the 100 amongst the remaining $x_{i}$ and $2^{9}$ ways to make any of them negative.So this gives you a total of $2^{9}\binom{10}{1}\binom{99}{8}$ ways. Case 3: Two out of $x_{i}$ is $0$: There are $\binom{10}{2}$ ways to make any two of $x_{i}$ equal to $0$ and $\binom{99}{7}$ ways to distribute the 100 amongst the remaining $x_{i}$ and $2^{8}$ ways to make any of them negative.So this gives you a total of $2^{8}\binom{10}{2}\binom{99}{7}$ ways. ................ Adding them up, you get the number of solutions: $$2^{10}\binom{99}{9}+2^{9}\binom{10}{1}\binom{99}{8}+2^{8}\binom{10}{2}\binom{99}{7}+2^{7}\binom{10}{3}\binom{99}{6}+2^{6}\binom{10}{4}\binom{99}{5}+2^{5}\binom{10}{5}\binom{99}{4}+2^{4}\binom{10}{6}\binom{99}{3}+2^{3}\binom{10}{7}\binom{99}{2}+2^{2}\binom{10}{8}\binom{99}{1}+2^{1}\binom{10}{9}$$.
Note: I can't edit ! :mad: : All signs in the expression above are positive ! In General, $$\left | x_{1} \right |+\left | x_{2} \right |+...+\left | x_{n} \right |= m$$Stars and bars give that there are $\binom{n}{k}\binom{m-1}{k-1}$ solutions where $k$ numbers are positive and the rest are zero. Each of these give rise to $2^{k}$ integer solutions. Hence, the total number of solutions is $$\sum_{k=1}^{n}2^{k}\binom{n}{k}\binom{m-1}{k-1}$$